# SOLUTION: Really stuck on this one! For the quadratic function y =F(x) = x^2-2x-15 i need the vertex and x & Y intercepts, graph, domain & range..thank you so much, Carol

Algebra ->  Algebra  -> Graphs -> SOLUTION: Really stuck on this one! For the quadratic function y =F(x) = x^2-2x-15 i need the vertex and x & Y intercepts, graph, domain & range..thank you so much, Carol      Log On

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 Question 58872: Really stuck on this one! For the quadratic function y =F(x) = x^2-2x-15 i need the vertex and x & Y intercepts, graph, domain & range..thank you so much, CarolAnswer by Edwin McCravy(8896)   (Show Source): You can put this solution on YOUR website!``` Really stuck on this one! For the quadratic function y = f(x) = x^2-2x-15 i need the vertex and x & Y intercepts, graph, domain & range..thank you so much, Carol The x-coordinate of the vertex of the parabola y = f(x) = ax² + bx + c is h = -b/(2a) And the y-coordinate of the vertex is k = f(h) Its domain is (-¥, ¥) and the range is [k, ¥) if a > 0 and (-i, k] if a < 0 The y-intercept is (0, c) The x intercepts are the (r1,0), (r2,0) where r1 and r2 are the solutions to the equation ax² + bx + c = 0 if they are real numbers. If they are imaginary numbers there are no x intercepts. Your problem is y = f(x) = x² - 2x - 15 so a = 1, b = -2, c = -15 The x-coordinate of the vertex of the parabola y = f(x) = x² - 2x - 15 is h = -b/(2a) = -(-2)/[2(1)] = 1 And the y-coordinate of the vertex is k = f(h) = f(1) = (1)² - 2(1) - 15 = -16 So the vertex is (h, k) = (-1, -16) Its domain is (-¥, ¥) and the range is [k, ¥) if a > 0 and since a = 1, which is > 0 and k = -16 the range is [-16, ¥ ) The y-intercept is (0, c) and since c = -15, the y-intercept is (0, -15) The x intercepts are (r1,0), (r2,0) where r1 and r2 are the solutions to the equation ax² + bx + c = 0 if they are real numbers. Solve this equation: x² - 2x - 15 = 0 you will get solutions -3 and 5 so the x-intercepts are (-3,0) and (5,0) The graph is: Edwin```